Answer
Diverges
Work Step by Step
Consider $a_n=\dfrac{1}{2 \sqrt n+\sqrt [3] n}$ and $b_n=\dfrac{1}{\sqrt n}$
Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{1}{2 \sqrt n+\sqrt [3] n}}{\dfrac{1}{\sqrt n}}$
Thus, we have $ =\lim\limits_{n \to \infty} \dfrac{\sqrt n}{2 \sqrt n+\sqrt [3] n}$
or, $=\dfrac{1}{2}$
Hence, the series diverges due to the limit comparison test.